Question: $h(t) = -t^{2}+g(t)$ $f(x) = -x$ $g(n) = -n+2(f(n))$ $ h(g(6)) = {?} $
Answer: First, let's solve for the value of the inner function, $g(6)$ . Then we'll know what to plug into the outer function. $g(6) = -6+2(f(6))$ To solve for the value of $g$ , we need to solve for the value of $f(6)$ $f(6) = -6$ $f(6) = -6$ That means $g(6) = -6+(2)(-6)$ $g(6) = -18$ Now we know that $g(6) = -18$ . Let's solve for $h(g(6))$ , which is $h(-18)$ $h(-18) = -(-18)^{2}+g(-18)$ To solve for the value of $h$ , we need to solve for the value of $g(-18)$ $g(-18) = -(-18)+2(f(-18))$ To solve for the value of $g$ , we need to solve for the value of $f(-18)$ $f(-18) = -(-18)$ $f(-18) = 18$ That means $g(-18) = -(-18)+(2)(18)$ $g(-18) = 54$ That means $h(-18) = -(-18)^{2}+54$ $h(-18) = -270$